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Question: If $\begin{bmatrix} 3 & 1 \\ 4 & 1 \end{bmatrix} X = \begin{bmatrix} 5 & -1 \\ 2 & 3 \end{bmatrix}$,...

If [3141]X=[5−123]\begin{bmatrix} 3 & 1 \\ 4 & 1 \end{bmatrix} X = \begin{bmatrix} 5 & -1 \\ 2 & 3 \end{bmatrix}, then X=X =

A

[−3414−13]\begin{bmatrix} -3 & 4 \\ 14 & -13 \end{bmatrix}

B

[3−4−1413]\begin{bmatrix} 3 & -4 \\ -14 & 13 \end{bmatrix}

C

[341413]\begin{bmatrix} 3 & 4 \\ 14 & 13 \end{bmatrix}

D

[−34−1413]\begin{bmatrix} -3 & 4 \\ -14 & 13 \end{bmatrix}

Answer

[−3414−13]\begin{bmatrix} -3 & 4 \\ 14 & -13 \end{bmatrix}

Explanation

Solution

We have [3141]X=[5−123]\begin{bmatrix} 3 & 1 \\ 4 & 1 \end{bmatrix} X = \begin{bmatrix} 5 & -1 \\ 2 & 3 \end{bmatrix}.

Let A=[3141]A = \begin{bmatrix} 3 & 1 \\ 4 & 1 \end{bmatrix} and Y=[5−123]Y = \begin{bmatrix} 5 & -1 \\ 2 & 3 \end{bmatrix}. Then, X=A−1YX = A^{-1}Y where det(A)=3⋅1−4⋅1=−1\text{det}(A) = 3\cdot1 - 4\cdot1 = -1.

The inverse of AA is given by: A−1=1−1[1−1−43]=[−114−3]A^{-1} = \frac{1}{-1}\begin{bmatrix} 1 & -1 \\ -4 & 3 \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ 4 & -3 \end{bmatrix}.

Now, multiply: X=[−114−3][5−123]X = \begin{bmatrix} -1 & 1 \\ 4 & -3 \end{bmatrix} \begin{bmatrix} 5 & -1 \\ 2 & 3 \end{bmatrix}.

Calculate each entry:

  • Top-left: (−1)(5)+1⋅2=−5+2=−3(-1)(5)+1\cdot2 = -5+2 = -3.
  • Top-right: (−1)(−1)+1⋅3=1+3=4(-1)(-1)+1\cdot3 = 1+3 = 4.
  • Bottom-left: 4(5)+(−3)(2)=20−6=144(5)+(-3)(2) = 20-6 = 14.
  • Bottom-right: 4(−1)+(−3)(3)=−4−9=−134(-1)+(-3)(3) = -4-9 = -13.

Thus, X=[−3414−13]X=\begin{bmatrix} -3 & 4 \\ 14 & -13 \end{bmatrix}.